Chapter 18, Problem 18.65QP

For each of the following redox reactions, (i) write the half-reactions, (ii) write a balanced equation for the whole reaction, (iii) determine in which direction the reaction will proceed spontaneously under standard-state conditions.

(a) H₂(g) + Ni²⁺(aq) → H⁺(aq) + Ni(s)

(b) MnO₄⁻(aq) + Cl⁻(aq) →     Mn²⁺(aq) + Cl₂(g)

(c) Cr(s) + Zn²⁺(aq) → Cr³⁺(aq) + Zn(s)

Interpretation Introduction

Interpretation:

For each of the given redox reactions, the half-cell reactions, the completely balanced cell reaction and the direction of spontaneous reactions has to be found.

Concept Introduction:

Redox reactions are the reactions in which both oxidation and reduction takes place simultaneously.  Oxidation is the removal electron from an atom or ion.  Oxidation process increases the oxidation number.  Reduction is the addition of electron to an atom or ion.  Reduction process decreases the oxidation number.  The electrochemical reaction of zinc with copper sulphate is an example of redox reaction.

${\text{Zn}}_{\text{(s)}}{\text{+ CuSO}}_{\text{4}}{}_{\text{(aq)}}\to {\text{ZnSO}}_{\text{4}}{\text{+ Cu}}_{\text{(s)}}$

Standard reduction potential is the measure of the tendency of a species to undergo reduction.  It is measured in terms of volts.  The substance which is having high positive value will easily undergo reduction.

The standard electrode potential of a cell ${\text{(E}}^{°}{}_{\text{cell}})$ is the difference in electrode potential of the cathode and anode.

${\text{E}}^{\text{°}}{}_{\text{cell}}={\text{E}}^{\text{°}}{}_{\text{cathode}}-{\text{E}}^{\text{°}}{}_{\text{anode}}$

The relation between Gibbs free energy and cell potential: The amount of energy in a system that can be converted into useful energy is defined as free energy in thermodynamics.

Free energy and the cell potential is related by the given equation.

$\text{ΔG=-nFE}$

Where,

$\text{ΔG}$ is the change in free energy

$\text{n}$ is the number of electrons transferred

$\text{F}$ is the Faraday constant $\text{(F}=\text{96485}{\text{.338Cmol}}^{\text{-1}})$

$\text{E}$ is the cell potential

Answer to Problem 18.65QP

(i)

The half-cell reactions are,

$\begin{array}{l}{\text{H}}_{\text{2}}\text{(g)}\to {\text{2H}}^{+}(\text{aq})+2e^{-}{\text{E}}^{°}=\text{}\text{0V}\\ {\text{Ni}}^{\text{2+}}(\text{aq})+{\text{2e}}^{\text{-}}\to \text{Ni(s)}{\text{E}}^{°}=-0.\text{25V}\end{array}$

(ii)

The completely balanced equation is,

${\text{H}}_{\text{2}}{\text{(g)+Ni}}^{\text{2+}}(\text{aq})\to {\text{2H}}^{+}(\text{aq})+\text{Ni(s)}$

(iii)

The reaction will be spontaneous towards to left side of the completely balanced equation.

Explanation of Solution

(i)

To write the half-cell reactions

The half-cell reactions for the given redox reactions are,

$\begin{array}{l}{\text{H}}_{\text{2}}{}_{\text{(g)}}\to {\text{2H}}^{+}{}_{(\text{aq})}+2e^{-}{\text{E}}^{°}=\text{}\text{0V}\\ {\text{Ni}}^{\text{2+}}{}_{(\text{aq})}+{\text{2e}}^{\text{-}}\to {\text{Ni}}_{\text{(s)}}{\text{E}}^{°}=-0.\text{25V}\end{array}$

(ii)

To write complete equation for the given redox reaction

The balanced equation for the given reaction can be represented as given below

${\text{H}}_{\text{2}}{}_{\text{(g)}}{\text{+ Ni}}^{\text{2+}}{}_{(\text{aq})}\to {\text{2H}}^{+}{}_{(\text{aq})}+{\text{Ni}}_{\text{(s)}}$

(iii)

To determine the direction of the spontaneous reaction in the given standard state.

In the electrochemical series the position of nickel is below the hydrogen.  Hence nickel will have the tendency to get oxidized.  The spontaneity of the reaction depends upon the change of the free energy.  Free energy and the electrode potential are related by the following equation.

$\text{ΔG=-nFE}$

In order to have a negative change in free energy the value of cell potential should be positive.

The cell potential of the given cell can be calculated by the following equation.

${\text{E}}^{\text{°}}{}_{\text{cell}}={\text{E}}^{\text{°}}{}_{\text{cathode}}-{\text{E}}^{\text{°}}{}_{\text{anode}}$

The one with higher positive value of reduction potential will be cathode and the one with lower value of reduction potential will be anode.  In the given reaction will be spontaneous when zinc is oxidised (anode) to ${\text{Zn}}^{\text{2+}}$ and hydrogen ion is reduced to hydrogen gas (cathode).  Hence, the spontaneous reaction will be the reverse of the overall reaction.  The reaction will be proceeding to the left side.

Interpretation Introduction

Interpretation:

For each of the given redox reactions, the half-cell reactions, the completely balanced cell reaction and the direction of spontaneous reactions has to be found.

Concept Introduction:

Redox reactions are the reactions in which both oxidation and reduction takes place simultaneously.  Oxidation is the removal electron from an atom or ion.  Oxidation process increases the oxidation number.  Reduction is the addition of electron to an atom or ion.  Reduction process decreases the oxidation number.  The electrochemical reaction of zinc with copper sulphate is an example of redox reaction.

${\text{Zn}}_{\text{(s)}}{\text{+ CuSO}}_{\text{4}}{}_{\text{(aq)}}\to {\text{ZnSO}}_{\text{4}}{\text{+ Cu}}_{\text{(s)}}$

Standard reduction potential is the measure of the tendency of a species to undergo reduction.  It is measured in terms of volts.  The substance which is having high positive value will easily undergo reduction.

The standard electrode potential of a cell ${\text{(E}}^{°}{}_{\text{cell}})$ is the difference in electrode potential of the cathode and anode.

${\text{E}}^{\text{°}}{}_{\text{cell}}={\text{E}}^{\text{°}}{}_{\text{cathode}}-{\text{E}}^{\text{°}}{}_{\text{anode}}$

The relation between Gibbs free energy and cell potential: The amount of energy in a system that can be converted into useful energy is defined as free energy in thermodynamics.

Free energy and the cell potential is related by the given equation.

$\text{ΔG=-nFE}$

Where,

$\text{ΔG}$ is the change in free energy

$\text{n}$ is the number of electrons transferred

$\text{F}$ is the Faraday constant $\text{(F}=\text{96485}{\text{.338Cmol}}^{\text{-1}})$

$\text{E}$ is the cell potential

Answer to Problem 18.65QP

(i)

The half-cell reactions are,

$\begin{array}{l}{\text{5e}}^{\text{-}}{\text{+8H}}^{\text{+}}{\text{(aq)+MnO}}_{\text{4}}{}^{\text{-}}\to {\text{Mn}}^{\text{2+}}{\text{(aq)+4H}}_{\text{2}}\text{O(l)}\\ {\text{2Cl}}^{-}(\text{aq)}\to {\text{Cl}}_2(g)+{\text{2e}}^{\text{-}}\end{array}$

(ii)

The completely balanced equation is,

$16{\text{H}}^{+}(\text{aq})+{\text{2MnO}}_4{}^{-}{\text{+10Cl}}^{-}\text{(aq)}\to {\text{2Mn}}^{4+}(\text{aq)}+8{\text{H}}_{\text{2}}{\text{O(l)+5Cl}}_{\text{2}}\text{(g)}$

(iii)

The reaction will be spontaneous towards to right side of the completely balanced equation.

Explanation of Solution

(i)

To write the half-cell reactions

The half-cell reactions for the given redox reactions are,

$\begin{array}{l}{\text{5e}}^{\text{-}}{\text{+8H}}^{\text{+}}{}_{\text{(aq)}}{\text{+MnO}}_{\text{4}}{}^{\text{-}}\to {\text{Mn}}^{\text{2+}}{}_{\text{(aq)}}{\text{+4H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{E}}^{\text{°}}{}_{\text{cell}}\text{= +1}\text{.51V}\\ {\text{2Cl}}^{-}{}_{(\text{aq)}}\to {\text{Cl}}_2{}_{(g)}+{\text{2e}}^{\text{-}}{\text{E}}^{\text{°}}{}_{\text{cell}}\text{= +1}\text{.36V}\end{array}$

(ii)

To write complete equation for the given redox reaction

The balanced equation for the given reaction can be represented as given below

$16{\text{H}}^{+}{}_{(\text{aq})}+{\text{2MnO}}_4{}^{-}{\text{+10Cl}}^{-}{}_{\text{(aq)}}\to {\text{2Mn}}^{4+}{}_{(\text{aq)}}+8{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{ +5Cl}}_{\text{2}}{}_{\text{(g)}}$

(iii)

To determine the direction of the spontaneous reaction in the given standard state.

In the electrochemical series the position of ${\mathrm{Cl}}^{-}(aq)$ is below the ${\text{MnO}}_4{}^{-}(aq)$.  Hence ${\mathrm{Cl}}^{-}{}_{(aq)}$ will have the tendency to get oxidized.  The spontaneity of the reaction depends upon the change of the free energy.  Free energy and the electrode potential are related by the following equation.

$\text{ΔG=-nFE}$

In order to have a negative change in free energy the value of cell potential should be positive.

The cell potential of the given cell can be calculated by the following equation.

${\text{E}}^{\text{°}}{}_{\text{cell}}={\text{E}}^{\text{°}}{}_{\text{cathode}}-{\text{E}}^{\text{°}}{}_{\text{anode}}$

The one with higher positive value of reduction potential will be cathode and the one with lower value of reduction potential will be anode.  In the given reaction will be spontaneous when the ${\mathrm{Cl}}^{-}{}_{(aq)}$ is oxidised ${\mathrm{Cl}}_2$ (anode) and ${\text{Mn}}^{7+}$ is reduced to ${\text{Mn}}^{2+}$ (cathode).  Hence, reaction will be proceeding to the right side.

Interpretation Introduction

Interpretation:

For each of the given redox reactions, the half-cell reactions, the completely balanced cell reaction and the direction of spontaneous reactions has to be found.

Concept Introduction:

Redox reactions are the reactions in which both oxidation and reduction takes place simultaneously.  Oxidation is the removal electron from an atom or ion.  Oxidation process increases the oxidation number.  Reduction is the addition of electron to an atom or ion.  Reduction process decreases the oxidation number.  The electrochemical reaction of zinc with copper sulphate is an example of redox reaction.

${\text{Zn}}_{\text{(s)}}{\text{+ CuSO}}_{\text{4}}{}_{\text{(aq)}}\to {\text{ZnSO}}_{\text{4}}{\text{+ Cu}}_{\text{(s)}}$

Standard reduction potential is the measure of the tendency of a species to undergo reduction.  It is measured in terms of volts.  The substance which is having high positive value will easily undergo reduction.

The standard electrode potential of a cell ${\text{(E}}^{°}{}_{\text{cell}})$ is the difference in electrode potential of the cathode and anode.

${\text{E}}^{\text{°}}{}_{\text{cell}}={\text{E}}^{\text{°}}{}_{\text{cathode}}-{\text{E}}^{\text{°}}{}_{\text{anode}}$

The relation between Gibbs free energy and cell potential: The amount of energy in a system that can be converted into useful energy is defined as free energy in thermodynamics.

Free energy and the cell potential is related by the given equation.

$\text{ΔG=-nFE}$

Where,

$\text{ΔG}$ is the change in free energy

$\text{n}$ is the number of electrons transferred

$\text{F}$ is the Faraday constant $\text{(F}=\text{96485}{\text{.338Cmol}}^{\text{-1}})$

$\text{E}$ is the cell potential

Answer to Problem 18.65QP

(i)

The half-cell reactions are,

$\begin{array}{l}\text{Cr(s)}\to {\text{Cr}}^{3+}{\text{(aq)+3e}}^{-}\\ {\text{Zn}}^{2+}({\text{aq)+2e}}^{\text{-}}\to \text{Zn(s)}\end{array}$

(ii)

The completely balanced equation is,

${\text{2Cr(s)+3Zn}}^{2+}(\text{aq)}\to 3{\text{Cr}}^{3+}\text{(aq)+3Zn(s)}$

(iii)

The reaction will be spontaneous towards to left side of the completely balanced equation.

Explanation of Solution

(i)

To write the half-cell reactions

The half-cell reactions for the given redox reactions are,

$\begin{array}{l}\text{Cr(s)}\to {\text{Cr}}^{3+}{\text{(aq)+3e}}^{-}{\text{E}}^{°}{}_{cell}=-0.74\text{V}\\ {\text{Zn}}^{2+}({\text{aq)+2e}}^{\text{-}}\to \text{Zn(s)}{\text{E}}^{°}{}_{cell}=-0.76\text{V}\end{array}$

(ii)

To write complete equation for the given redox reaction

The balanced equation for the given reaction can be represented as given below

${\text{2Cr}}_{\text{(s)}}{\text{ + 3Zn}}^{2+}{}_{(\text{aq)}}\to 3{\text{Cr}}^{3+}{}_{\text{(aq)}}{\text{ + 3Zn}}_{\text{(s)}}$

(iii)

To determine the direction of the spontaneous reaction in the given standard state.

In the electrochemical series the position of $\text{Zn(s)}$ is below the ${\text{Cr}}^{3+}(aq)$.  Hence ${\text{Zn}}_{\text{(s)}}$ will have the tendency to get oxidized.  The spontaneity of the reaction depends upon the change of the free energy.  Free energy and the electrode potential are related by the following equation.

$\text{ΔG=-nFE}$

In order to have a negative change in free energy the value of cell potential should be positive.

The cell potential of the given cell can be calculated by the following equation.

${\text{E}}^{\text{°}}{}_{\text{cell}}={\text{E}}^{\text{°}}{}_{\text{cathode}}-{\text{E}}^{\text{°}}{}_{\text{anode}}$

The one with higher positive value of reduction potential will be cathode and the one with lower value of reduction potential will be anode.  In the given reaction will be spontaneous when the ${\text{Zn}}_{\text{(s)}}$ is oxidised ${\text{Zn}}^{2+}$ (anode) and ${\text{Cr}}^{3+}$ is reduced to $\text{Cr(s)}$ (cathode).  Hence, reaction will be proceeding to the left side.